Complex analysis proof with maximum $f(z)$ is analytic in {$z:|z|>1/2$}.
$|f(z)|\le1$ for every $|z|=1$ and $\lim_{z\to\infty}\frac{f(z)}{z^3}$ it final.
Prove that $|f(z)|\le|z|^3$ for every $|z|\ge1$.
After that, if in addion there is $|z_0|>1$ so that $f(z_0)=z_0^3$, then prove $f(z)=z^3$ for every $|z|>1/2$.
 A: This problem can probably be solved as stated by working in $\hat {\Bbb{C}}$ but it's easier (at least for me) to make an inversion between $0$ and complex infinity and solve the problem in a disk around the origin.
Define
$$g(z) := z^3 f\left(\frac 1 z\right)$$
We know that $g$ is analytic in the punctured disk $0 < |z| < 2$ (since $f$ is analytic in $|z| > 1/2$). Now,
$$\lim_{z\to 0} g(z) = \lim_{z\to 0} z^3 f\left(\frac 1 z\right) = \lim_{z\to \infty} \frac {f(z)} {z^3} = \text{(finite)}$$
By Riemann's Removable Singularity Theorem, we can redefine $g$ at $z=0$ in such a way that $g$ is analytic in the disk $|z| < 2$ including at $z=0$.
Now, since whenever $|z| = 1$, we also have $|1/z| = 1$ and $|z^3| = 1$, we have that $|g(z)| \le 1$ on the circle $|z|=1$.
By the "global" version of the maximum modulus principle, we also have that $|g(z)| \le 1$ whenever $|z| \le 1$. So:
$$|z^3| \left| f \left( \frac 1 z \right) \right| \le 1 \ \ \ \ \ \forall|z|\le 1$$
Replacing $z$ with $1/z$ gives the wanted inequality.
Finally, if there is some $|z_0|>1$  for which $f(z_0) = z_0^3$, then $z_1=1/z_0$ satisfies $g(z_1) = 1$. Again by the maximum modulus principle, this requires $g$ to be constant in the disk $|z| < 1$ (and the constant must be the value at $z_1$, i.e. $1$). By using the definition of $g$ we get that $f(z) = z^3$ when $|z| > 1$. This may be immediately extended to $|z| > 1/2$ by analyticity.
